Find the equation of the line tangent to the curve $y=x^2$ parallel to the line $y=x$

Find the equation of the line tangent to the curve $y=x^2$ parallel to the line $y=x$.

Just started A level maths, any help is appreciated.

• The slope of the line $y=x$ is $1$. That should be obvious. We must find where the slope of the curve $y=x^2$ is equal to $1$. Do you know how to find the slope of a curve? – Emily Oct 16 '14 at 19:13
• Yes, I know how to find the slope of a curve. Thanks – Joe Oct 16 '14 at 19:15
• Ok, so what is the equation that describes the slope of the curve $y=x^2$? – Emily Oct 16 '14 at 19:22

The slope of $y=x$ is $1$.

We ask ourselves the following question: in what point is the tangent at $y=x^2$ $1$?

$y'=2x=1$ gives $x=\frac12$.

$y(\frac12)=\frac14$, so the line passing through $(x,y)=(\frac12,\frac14)$ with slope $1$ is $y=x-\frac14$.

• Thanks very much, helped a lot! Does this method apply to all questions of this type? – Joe Oct 16 '14 at 19:16
• Yes, it applies to all questions of this type. – rae306 Oct 16 '14 at 19:17
• Thank you very much for your help. – Joe Oct 16 '14 at 19:19

Find $y'(x)$. When does $y'(x) = 1\;?$ ($m = 1$ is the slope of the line $y = x$, and hence the slope of any line parallel to $y = x$).

There will be one solution: name it $x_0$.

Find $y$ at $x_0$, and call it $y_0$.

Then you have the slope of the desired line, and the point $(x_0, y_0)$ and can use the point slope form of the equation of a line: $$y - y_0 = m(x-x_0)$$